272 Wilder’s Algebraic Solution. 
3 
xri+z3— sp2+4; =c, (3). These three 
equations son to x«?-+(y+z)e+p=0, (D), are sufficient 
to determ 
For Srevity ‘tsk z=0, then 
(2) p=-% 
and (2)and(3) z?= 5 Van =¢(bc) 
2 
forthen. (D) y=-— “tte mena 2 ) becomes 
_@tp) which is the rule of 
Cardan. . 
4 2 Ss, 
So also the function ane — ee = st ’ a , be- 
comes, on Sn is a factor of (B) indgpendentiy of z, 
. 
aa a 
yx? +-pxr 
writing (2ny+2z) fr Ys aad (aetys Tet 2z2—p) for 
p, and . 4 change 
tigre. says y*+16n°® a z*— 4n*p)y? + 
+(2ny +2z)a? + 
(16nz* — 8npz —4nq)y +424 — 4pz* +p? sere mee 
(2n2y? +4nzy +22? —p)x r 
dividing by 4n‘*z?, the a (E)=0, and — comparing 
with y* y > + by? +-cy-+-d= =o 
ed 
we shall have or ey (1) 
~(62*p)=b, (2) 
1 
seca (3) 
1 xr BY oats 7" 
pore A +2z'—p 1224 gz Lym, (4). 
These equations joined to 
22+ (2ny-+2z)0% H(2n?y? +4nz Aan peli He ee 
are sufficient to determine y. =0 and n=!, 
=a xe +2bxt+(b? —4d)z? Atoms each; is a ainaae 
of Des Cartes. When z=0 and n=1, we have 
